The question is about a particular term of an Arithmetic Progression. It asks about 'x' times of the xth term. With some simple but very powerful ideas, one can cut down on a lot of working when it comes to progressions. For example, anchoring a progression around its middle term can be very useful. CAT Exam tests the idea of progression often in the CAT Quantitative Aptitude section and this could also be tested in DI LR section of the CAT Exam as a part of a puzzle.
Question 7: If 4 times the 4th term of an A.P. is equal to 9 times the 9th term of the A.P., what is 13 times the 13th term of this A.P.?
4t4 = 9t9 , we need to find t13
4(a + 3d) = 9(a + 8d)
4a + 12d = 9a + 72d
=> 5a + 60d = 0
=> a + 12d = 0
=> t13 = 0
=> 13 * t13 = 0
As a simple rule, if n * tn = m * tm, then tm+n = 0. See, if you can prove this.
The question is "what is 13 times the 13th term of this A.P.?"
Choice B is the correct answer.
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